MARKSCHEME - ednet.ns.cahrsbstaff.ednet.ns.ca/woodsa/IB year 2/Mathematical methods SL... · PROGRAMA DEL DIPLOMA DEL BI M05/5/MATME/SP2/ENG/TZ0/XX/M+ MARKSCHEME May 2005 ... Examples

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IB DIPLOMA PROGRAMME PROGRAMME DU DIPLÔME DU BI PROGRAMA DEL DIPLOMA DEL BI

M05/5/MATME/SP2/ENG/TZ0/XX/M+

MARKSCHEME

May 2005

MATHEMATICAL METHODS

Standard Level

Paper 2 21 pages

M05/5/MATME/SP2/ENG/TZ0/XX/M+ – 2 –

This markscheme is confidential and for the exclusive use of examiners in this examination session. It is the property of the International Baccalaureate and must not be reproduced or distributed to any other person without the authorization of IBCA.


Instructions to Examiners Note: Where there are two marks (e.g. M2, A2) for an answer do not split the marks unless

otherwise instructed. 1 Method of marking (a) All marking must be done using a red pen. (b) Marks should be noted on candidates’ scripts as in the markscheme:

show the breakdown of individual marks using the abbreviations (M1), (A2) etc., unless a part is completely correct; write down each part mark total, indicated on the markscheme (for example, [3 marks] ) – it

is suggested that this be written at the end of each part, and underlined; write down and circle the total for each question at the end of the question.

2 Abbreviations The markscheme may make use of the following abbreviations: (M) Marks awarded for Method (A) Marks awarded for an Answer or for Accuracy (N) Marks awarded for correct answers, if no working (or no relevant working) shown: they may

not necessarily be all the marks for the question. Examiners should only award these marks for correct answers where there is no working.

(R) Marks awarded for clear Reasoning (AG) Answer Given in the question and consequently marks are not awarded

Note: Unless otherwise stated, it is not possible to award (M0)(A1). Follow through (ft) marks should be awarded where a correct method has been attempted but error(s)

are made in subsequent working which is essentially correct. • Penalize the error when it first occurs • Accept the incorrect result as the appropriate quantity in all subsequent working • If the question becomes much simpler then use discretion to award fewer marks

Examiners should use (d) to indicate where discretion has been used. It should only be used for

decisions on follow through and alternative methods. It must be accompanied by a brief note to explain the decision made

3 Using the Markscheme (a) This markscheme presents a particular way in which each question may be worked and how it

should be marked. Alternative methods have not always been included. Thus, if an answer is wrong then the working must be carefully analysed in order that marks are awarded for a different method in a manner which is consistent with the markscheme. Indicate the awarding of these marks by (d).


Where alternative methods for complete questions or parts of questions are included, they are indicated by METHOD 1, METHOD 2, etc. Other alternative part solutions are indicated by EITHER….OR. It should be noted that G marks have been removed, and GDC solutions will not be indicated using the OR notation as on previous markschemes.

Candidates are expected to show working on this paper, and examiners should not award full

marks for just the correct answer. Where it is appropriate to award marks for correct answers with no working (or no relevant working), it will be shown on the markscheme using the N notation. All examiners will be expected to award marks accordingly in these situations.

(b) Unless the question specifies otherwise, accept equivalent forms. For example: sincos

θθ

for tanθ .

On the markscheme, these equivalent numerical or algebraic forms will generally be written in brackets after the required answer. Paper setters will indicate the required answer, by allocating full marks at that point. Further working should be ignored, even if it is incorrect. For example: if candidates are asked to factorize a quadratic expression, and they do so correctly, they are awarded full marks. If they then continue and find the roots of the corresponding equation, do not penalize, even if those roots are incorrect, i.e. once the correct answer is seen, ignore further working.

(c) As this is an international examination, all alternative forms of notation should be accepted. For

example: 1.7 , 1·7, 1,7; different forms of vector notation such as 1, , ; tanu u u x− for arctan x. 4 Accuracy of Answers

If the level of accuracy is specified in the question, a mark will be allocated for giving the answer to

the required accuracy.

There are two types of accuracy error. Candidates should be penalized once only IN THE PAPER for an accuracy error (AP). Award the marks as usual then write –1(AP) against the answer and also on the front cover Rounding errors: only applies to final answers not to intermediate steps. Level of accuracy: when this is not specified in the question the general rule unless otherwise stated in the question all numerical answers must be given exactly or to three significant figures applies. • If a final correct answer is incorrectly rounded, apply the AP

OR • If the level of accuracy is not specified in the question, apply the AP for answers not given to 3

significant figures. (Please note that this has changed from 2003). Note: If there is no working shown, and answers are given to the correct two significant figures, apply

the AP. However, do not accept answers to one significant figure without working. 5 Graphic Display Calculators

Many candidates will be obtaining solutions directly from their calculators, often without showing any working. They have been advised that they must use mathematical notation, not calculator commands when explaining what they are doing. Incorrect answers without working will receive no marks. However, if there is written evidence of using a graphic display calculator correctly, method marks may be awarded. Where possible, examples will be provided to guide examiners in awarding these method marks.


Examples 1. Accuracy A question leads to the answer 4.6789….

• 4.68 is the correct 3 s.f. answer. • 4.7, 4.679 are to the wrong level of accuracy : both should be penalised the first time this type

of error occurs. • 4.67 is incorrectly rounded – penalise on the first occurrence.

Note: All these “incorrect” answers may be assumed to come from 4.6789..., even if that value is not seen, but previous correct working is shown. However, 4.60 is wrong, as is 4.5, 4.8, and these should be penalized as being incorrect answers, not as examples of accuracy errors.

2. Alternative solutions The points P, Q, R are three markers on level ground, joined by straight paths PQ, QR, PR as shown in

the diagram. ˆ ˆQR km, PQR , PRQ= 9 = 35 = 25 . (Note: in the original question, the first part was to find PR = 5.96) (a) Tom sets out to walk from Q to P at a steady speed of 8 km h−1 . At the same time,

Alan sets out to jog from R to P at a steady speed of km h .a −1 They reach P at the same time. Calculate the value of a. [7 marks]

(b) The point S is on [PQ], such that RS QS,= 2 as shown in the diagram. P S

Q R Find the length QS. [6 marks]

diagram not to scale

35 25 P

Q R9 km


MARKSCHEME (a) EITHER

Sine rule to find PQ

9sin 25PQsin120

= (M1)(A1)

PQ 4.39 km= (A1)

OR

Cosine rule: 2 2PQ 5.96 cos 252= + 9 − (2)(5.96)(9) (M1)(A1) 19.29= PQ 4.39 km= (A1)

THEN

Time for Tom 4.39=

8 (A1)

Time for Alana

5.96= (A1)

Then 4.398 a

5.96= (M1)

10.9a = (A1) (N5)

[7 marks] Note that the THEN part follows both EITHER and OR solutions, and this is shown by the alignment. (b) METHOD 1

2 2RS 4QS= (A1) 2 24QS QS QS cos35= + 81−18 × × (M1)(A1) 2 23QS 14.74QS 81 0 (or 3 14.74 81 0)x x+ − = + − = (A1) QS 8.20 or QS 3.29⇒ = − = (A1) therefore QS = 3.29 (A1)

METHOD 2

QS 2QSˆ sin35sinSRQ

= (M1)

1ˆsinSRQ sin 352

⇒ = (A1)

ˆSRQ 16.7= (A1) Therefore, ˆQSR 180 (35 16.7)= − + 128.3= (A1)

9 QS SRsin128.3 sin16.7 sin35

⎛ ⎞= =⎜ ⎟⎝ ⎠

(M1)

9sin16.7 9sin35QSsin128.3 2sin128.3

⎛ ⎞= =⎜ ⎟⎝ ⎠

3.29= (A1) (N2)

If candidates have shown no working, award (N5) for the correct answer 10.9 in part (a),

and (N2) for the correct answer 3.29 in part (b). [6 marks]


3. Follow through Question Calculate the acute angle between the lines with equations

4 41 3

s⎛ ⎞ ⎛ ⎞

= +⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠r and

2 14 1

t⎛ ⎞ ⎛ ⎞

= +⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠r .

Markscheme Angle between lines = angle between direction vectors (may be implied) (A1)

Direction vectors are 43⎛ ⎞⎜ ⎟⎝ ⎠

and 11

⎛ ⎞⎜ ⎟−⎝ ⎠

(may be implied) (A1)

4 1 4 1

cos3 1 3 1

θ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞

=⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠i (M1)

4 1 3 ( 1)× + × − = ( ) ( )( )22 2 24 3 1 1 cosθ+ + − (A1)

1cos5 2

θ = (= 0.1414….) (A1)

81.9θ = (1.43 radians) (A1) (N3) Examples of solutions and marking Solutions Marks allocated 1. 4 1 4 1

cos3 1 3 1

7cos5 28.13

θ

θ

θ

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

=

=

i

(A1)(A1) implied (M1) (A0)(A1) (A1)ft

Total 5 marks

2.

cosθ

4 21 4

17 20

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠=i

(A0)(A0)) wrong vectors implied (M1) for correct method, (A1)ft

0.2169= (A1)ft 77.5θ = (A1)ft Total 4 marks 3. 81.9θ = (N3) Total 3 marks

Note that this candidate has obtained the correct answer, but not shown any working. The way the markscheme is written means that the first 2 marks may be implied by subsequent correct working, but the other marks are only awarded if the relevant working is seen. Thus award the first 2 implied marks, plus the final mark for the correct answer.

END OF EXAMPLES


QUESTION 1

(a) (i) 2p = − 4q = (or 4, 2p q= = − ) (A1)(A1) (N1)(N1)

(ii) ( 2)( 4)y a x x= + − 8 (6 2)(6 4)a= + − (M1) 8 16a=

12

a = (A1) (N1)

(iii) 1 ( 2)( 4)2

y x x= + −

21 ( 2 8)2

y x x= − −

21 42

y x x= − − (A1) (N1)

[5 marks]

(b) (i) d 1dy xx

= − (A1) (N1)

(ii) 1 7x − = (M1)

( )8, 20 P is (8, 20)x y= = (A1)(A1) (N2) [4 marks]

(c) (i) when x = 4, gradient of tangent is 4 – 1 = 3 (may be implied) (A1)

gradient of normal is 13

− (A1)

1 1 40 ( 4)3 3 3

y x y x⎛ ⎞− = − − = − +⎜ ⎟⎝ ⎠

(A1) (N3)

(ii) 21 1 442 3 3

x x x− − = − + (or sketch/graph) (M1)

21 2 16 02 3 3

x x− − =

23 4 32 0x x− − = (may be implied) (A1) (3 8)( 4) 0x x+ − =

8 or 43

x x= − =

8 ( 2.67)3

x = − − (A1) (N2)

[6 marks]

Total [15 marks]


QUESTION 2

(a) (i) 80 8P( ) 0.381210 21

A ⎛ ⎞= = =⎜ ⎟⎝ ⎠

(A1) (N1)

(ii) 35 1P(year 2 art) 0.167210 6

⎛ ⎞= = =⎜ ⎟⎝ ⎠

(A1) (N1)

(iii) No (the events are not independent, or, they are dependent) (A1) (N1)

EITHER

P( ) P( ) P( )A B A B∩ = × (to be independent) (M1)

100 10P( ) 0.476210 21

B ⎛ ⎞= = =⎜ ⎟⎝ ⎠

(A1)

1 8 106 21 21≠ × (A1)

OR

P( )=P( )A A B (to be independent) (M1)

35P( )100

A B = (A1)

8 3521 100

≠ (A1)

OR

P( )=P( )B B A (to be independent) (M1)

100 10P( ) 0.476210 21

B ⎛ ⎞= = =⎜ ⎟⎝ ⎠

, 35P( )80

B A = (A1)

35 10080 210

≠ (A1)

Note: Award the first (M1) only for a mathematical interpretation of independence.

[6 marks] (b) (history) 85n = (A1)

50 10P(year 1 history) 0.58885 17

⎛ ⎞= = =⎜ ⎟⎝ ⎠

(A1) (N2)

[2 marks]

(c) 110 100 100 110 110 1002210 209 210 209 210 209

⎛ ⎞ ⎛ ⎞ ⎛ ⎞× + × = × ×⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

(M1)(A1)(A1)

( )200 0.501399

= = (A1) (N2)

[4 marks]

Total [12 marks]


QUESTION 3

(a) for using cosine rule ( )2 2 2 2 cosa b c ab C= + − (M1)

( )( )2 2 2BC 15 17 2 15 17 cos 29= + − (A1) BC 8.24 m= (A1) (N0)

Notes: Either the first or the second line may be implied, but not both. Award no marks if 8.24 is obtained by assuming a right (angled) triangle (BC=17sin29).

[3 marks]

(b) (i) A C 29

17 85

A CB 180 (29 85) 66= − + = for using sine rule (may be implied) (M1)

AC 17sin85 sin66

= (A1)

17sin85ACsin 66

=

AC (18.5380 ) 18.5 m= =… (A1) (N2)

(ii) ( )( )1Area 17 18.538... sin 292

= (A1)

276.4 m= (Accept 276.2 m ) (A1) (N1) [5 marks]

(c) ACB from previous triangle 66= Therefore alternative ˆACB 180 66 114 (or 29 85)= − = + (A1)

ˆABC 180 (29 114) 37= − + =

A C 29 114

17

37

B

AC 17sin37 sin114

= (M1)(A1)

AC (11.19906 ) 11.2 m= =… (A1) (N1) [4 marks]

continued…

B


Question 3 continued (d) A C 29 17

Minimum length for BC when ACB 90= or diagram showing right triangle (M1)

CBsin 2917

=

CB 17sin 29= CB (8.2417 ) 8.24 m= =… (A1) (N1)

[2 marks]

Total [14 marks]

B


QUESTION 4

(a) (i) 1( ) 2cos2 sin2

f x x x′ = × −

cos2 sinx x= − (A1)(A1) (N2)

Note: Award (A1)(A1) for 22sin sin 1x x− − + only if work shown, using product rule on sin cos cosx x x+ . (ii) 22sin sin 1x x+ − (2sin 1)(sin 1)x x= − + or 2(sin 0.5) (sin 1)x x− + (A1) (N1) (iii) 2sin 1 or sin 1x x= = −

1sin2

x =

π 5π 3π(0.524) (2.62) (4.71)6 6 2

x x x= = = = = = (A1)(A1)(A1) (N1)(N1)(N1)

[6 marks]

(b) ( )π 0.5246

x = = (A1) (N1)

[1 mark] (c) (i) EITHER

curve crosses axis when π2

x = (may be implied) (A1)

π 5π2 6π π6 2

Area ( )d ( )df x x f x x= +∫ ∫ (M1)(A1) (N3)

OR

Area =56

6

( ) df x xπ

π∫ (M1)(A2) (N3)

(ii) Area 0.875 0.875= + (M1) 1.75= (A1) (N2)

[5 marks]

Total [12 marks]


QUESTION 5

(a) (i) 200 600

AB400 200

→ −⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

(A1)

800600⎛ ⎞

= ⎜ ⎟⎝ ⎠

(A1) (N2)

(ii) 2 2AB 800 600 1000→

⎜ ⎟ = + = (must be seen) (M1)

unit vector 80016001000⎛ ⎞

= ⎜ ⎟⎝ ⎠

(A1)

0.80.6⎛ ⎞

= ⎜ ⎟⎝ ⎠

(AG) (N0)

Note: A reverse method is not acceptable in “show that” questions. [4 marks]

(b) (i) 0.8

2500.6⎛ ⎞

= ⎜ ⎟⎝ ⎠

v (M1)

200150⎛ ⎞

= ⎜ ⎟⎝ ⎠

(AG) (N0)

Note: A correct alternative method is using the given vector equation with t = 4.

(ii) at 13:00, t = 1

600 200

1200 150

xy

−⎛ ⎞ ⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠

(M1)

40050

−⎛ ⎞= ⎜ ⎟−⎝ ⎠

(A1) (N1)

(iii) AB 1000→

⎜ ⎟ =

Time 1000 4 (hours)250

= = (M1)(A1)

over town B at 16:00 (4 pm, 4:00 pm) (Do not accept 16 or 4:00 or 4) (A1) (N3)

[6 marks]

continued …


Question 5 continued (c) Note: There are a variety of approaches. The table shows some of them, with the mark allocation. Use discretion, following this allocation as closely as possible.

Time for A to B to C = 9 hours

Distance from A to B to C = 2250 km

Fuel used from A to B = 1800 4 7200× = litres (A1)

Light goes on after 16 000 litres

Light goes on after 16000 litres

Fuel remaining = 9800 litres (A1)

Time for 16 000 litres

16 0001800

=

88 ( 8.889)9

= =

Time remaining is

( )1 0.1119

= = hour

Distance on 16000 litres

16 000 2501800

= ×

( )22222 2222.229

= = km

Hours before light 88001800

( )84 4.8899

= =

Time remaining is

( )1 0.1119

= = hour

(A1)(A1)

(A1)

Distance 1 2509

= ×

= 27.8 km

Distance to C = 2250 – 2222.22

= 27.8 km

Distance 1 2509

= ×

= 27.8 km (A2)(N4)

[7 marks]

Total [17 marks]


QUESTION 6

(i) (a) 2 ( )xy

x

sy y x x

s− = − (M1)

23660 ( 10)3

y x− = − (A1)(A1)

4 20y x⇒ = + (A1) (N2) [4 marks]

(b) (i) 20 4 20 20x y= ⇒ = × + (M1) 100= (A1) (N0)

(ii) (a) 363(15)

xy

x y

sr

s s= = (M1)

= 0.8 (A1) (N1)

(b) The value of r indicates fairly good, (or equivalent) (R1)

but not exceptionally (moderately, fairly) strong (R1) linear correlation. Thus, x = 20 does not guarantee y = 100 (is not reliable) (A1) (Use discretion for equivalent meanings.)

[7 marks] (ii) (a) (i) P ( 3 200) P( 0.4)X Z> = > (M1) 1 0.6554 34.5 % ( 0.345)= − = = (A1) (N2) (ii) P (2 300 3 300)X< < = P( 1.4 0.6)Z− < < (M1) 0.4192 0.2257= + 0.645= (A1) 2P (both) (0.645) 0.416= = (A1) (N2) (iii) 0.7422 P( 0.65)Z= < (A1)

3 000 0.65500

d −= (A1)

$ 3 325 ( $ 3 330 to 3 s.f.)d = = (Accept $ 3325.07) (A1) (N3) [8 marks]

continued…


Question 6(ii) continued (b) (i) 0H : The mean savings is $ 3000 .( 0H : 3000µ = ) (A1) (N1) 1H : The mean savings is less than $ 3000 .( 1H : 3000µ < ) (A1) (N1)

(ii) METHOD 1

S.E. 500 50100

= = (A1)

Thus 2950 3000 150

z −= = − (A1)

EITHER The critical region for a one-tailed test at the 5 % level is 1.645z < − ( or 1.645z > ) (R1) Since z is not in the critical region (i.e. 1 1.645− > − )(or 1 < 1.645), we accept the null hypothesis. (or fail to reject) (R1) Therefore, there is not sufficient evidence to support the newspaper’s claim. (AG) (N0)

OR

( 1) 0.1587p z < − = (A1) 0.1587 0.05> (R1) so we accept the null hypothesis (AG) (N0)

METHOD 2

0.1587p = (A2) Since 0.1587 0.05> , (R1) we accept the null hypothesis. (or fail to reject) (R1) Therefore, there is not sufficient evidence to support the newspaper’s claim. (AG) (N0) (iii) 2950 2.58(50) 2950 2.58(50)µ− < < + (M1)(A1)(A1)

2.58 50 129× = 99 % C.I. is 2950 129 (2821, 3079) or (2820, 3080)± (A1)(A1) (N5) [11 marks]

Total [30 marks]


QUESTION 7

(i) (a) 15

x = or 5 1 0x − = (A1) (N1)

[1 mark]

(b) 2

2

(5 1)(6 ) (3 )(5)( )(5 1)

x x xf xx

− −′ =−

(M1)(A1)

2 2

2

30 6 15(5 1)

x x xx− −

=−

(may be implied) (A1)

2

2

15 6(5 1)

x xx−

=−

(accept a = 15, b = –6) (A1) (N2)

[4 marks] (ii) (a) x = 1 (A1)

EITHER

The gradient of ( )g x goes from positive to negative (R1)

OR

( )g x goes from increasing to decreasing (R1)

OR

when 1, ( )x g x′′= is negative (R1) [2 marks]

(b) 3 2 1 3x x− < < − < <and (A1) ( )g x′ is negative (R1)

[2 marks]

(c) 12

x = − (A1)

EITHER

( )g x′′ changes from positive to negative (R1)

OR

concavity changes (R1) [2 marks]

continued…


Question 7 (ii) continued (d)

(A3) [3 marks]

(iii) (a) 2 1u x= + , d 2 du x x= (A1)

26 1dx x x+∫123 du u= ∫ (A1)

3223

3u c= × +

322u c= + (A1)

= 3

2 22( 1)x c= + + (A1) (N4)

[4 marks]

(b) 3 2

06 1dx x x+∫

332 2

0

2( 1)x⎡ ⎤

= +⎢ ⎥⎣ ⎦

(A1)

3 32 22(4) 2(1)= − 16 2= −

14= (A1) (N2) [2 marks]

(c) ( )0

14 2 5 dk

x x= +∫ (M1)

2

014 5

kx x⎡ ⎤= +⎣ ⎦ (A1)

214 5k k= + ( )20 5 14k k= + − 0 ( 2)( 7)k k= − + 2 or 7k k= = − (A1)(A1) (N1)(N1)

[4 marks]

continued…


Question 7 continued

(iv) (a)

(i) Inscribed trapeziums (see above) (A2)

(ii) 12

h = (may be implied) (A1)

for correct setup of trapezium rule (A1)

[ ]1 1 (2) 2 (2.5) 2 (3) 2 (3.5) (4)2 2

A f f f f f⎛ ⎞= + + + +⎜ ⎟⎝ ⎠

0.25[5 13.5 16 17.5 9]= + + + + 15.25 ( 15.3)= = (A1) (N2)

[5 marks]

(b) EITHER It underestimates the actual area since there is some area that is under the curve

but is not included in any of the trapeziums (could be shown by a diagram) (R1)

OR the curve is concave down (R1)

[1 mark]

Total [30 marks]


QUESTION 8

(i) (a) 4 8 5 2

32 14 1 a−⎛ ⎞ ⎛ ⎞

= −⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠Q (A1)

9 6

33 14 a−⎛ ⎞

= ⎜ ⎟−⎝ ⎠Q (A1)

3 2

1413

a−⎛ ⎞⎜ ⎟= −⎜ ⎟⎜ ⎟⎝ ⎠

Q (A1) (N3)

[3 marks]

(b) 2 4 5 2

1 7 1 a−⎛ ⎞ ⎛ ⎞

= ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠CD

14 4 42 2 7

aa

− − +⎛ ⎞= ⎜ ⎟− +⎝ ⎠

(A1)(A1)(A1)(A1) (N4)

[4 marks] (c) det 5 2a= +D (may be implied) (A1)

1 211 55 2a

a− −⎛ ⎞

= ⎜ ⎟+ ⎝ ⎠D (A1) (N2)

[2 marks]

(ii) (a) (i) 0

0k

k⎛ ⎞

= ⎜ ⎟⎝ ⎠

E (A1)

0 1 4

0 2 8k

k− −⎛ ⎞⎛ ⎞ ⎛ ⎞

=⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎝ ⎠⎝ ⎠ ⎝ ⎠

4 00 4⎛ ⎞

= ⎜ ⎟⎝ ⎠

E (A1) (N2)

(ii) cos sinsin cos

α αα α

−⎛ ⎞= ⎜ ⎟⎝ ⎠

R

cos sin 5 4sin cos 0 3

α αα α

−⎛ ⎞⎛ ⎞ ⎛ ⎞=⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠ (A1)

4 35 53 45 5

⎛ ⎞−⎜ ⎟= ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

R (A1)(A1) (N2)

continued…


Question 8(ii)(a) continued

(iii) 10 1

k⎛ ⎞= ⎜ ⎟⎝ ⎠

H (A1)

1 5 200 1 5 5

k⎛ ⎞⎛ ⎞ ⎛ ⎞=⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠

5 5 20k+ = (A1) 3k = (A1)

1 30 1⎛ ⎞

= ⎜ ⎟⎝ ⎠

H (A1) (N4)

[9 marks] (b) det 0≠H , or a geometric reason, or the inverse could be found. (R1)

[1 mark]

(iii) (a) 2 1

(2 5 ) 2 51 3

−⎛ ⎞ ⎛ ⎞− = −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠M v w (M1)

13h = − (A1) (N2) [2 marks]

(b) 1 2 5 20 1 2 1 2 1

x xx x

+⎛ ⎞⎛ ⎞ ⎛ ⎞=⎜ ⎟⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠ ⎝ ⎠

5 2

or2 1

x ty t

= +⎛ ⎞⎜ ⎟= +⎝ ⎠

(M1)

2 1or5 2

x yt t− −= = (A1)

2 15 5

y x= + (A1) (N1)

Note: An alternate method is to find two correct points, their images and then the line. Award (A1) for each correct image and (A1) for the line.

[3 marks] (c) (i) 1−=W T S (M1)

1 1 20 1

− −⎛ ⎞= ⎜ ⎟⎝ ⎠

T (A1)

3 6

1 4− −⎛ ⎞

= ⎜ ⎟⎝ ⎠

W (A1) (N2)

(ii) det 6 or det 6= − =S S (A1) Area of 1A = (A1) Area of 6 1 6A′ = × = (A1) (N2)

[6 marks]

Total [30 marks]

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MARKSCHEME - ednet.ns.cahrsbstaff.ednet.ns.ca/woodsa/IB year 2/Mathematical methods SL... · PROGRAMA DEL DIPLOMA DEL BI M05/5/MATME/SP2/ENG/TZ0/XX/M+ MARKSCHEME May 2005 ... Examples